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Exercise 2.5 in Vijay Krishna's Auction Theory (slightly modified):

Consider a two-bidder first-price auction in which bidders’ values are distributed according to $F$. Let $\beta$ be the symmetric equilibrium, i.e., $$\beta(x)=E[Y\mid Y<x],$$ where $Y$ is drawn from $F$.

Now suppose that after the auction is over, both the losing and winning bids are publicly announced. In addition, there is the possibility of postauction resale: The winner of the auction may, if he so wishes, offer the object to the other bidder at a fixed “take-it-or-leave-it” price of $p$. If the other bidder agrees, then the object changes hands, and the losing bidder pays the winning bidder $p$. Otherwise, the object stays with the winning bidder, and no money changes hands. The possibility of postauction resale in this manner is commonly known to both bidders prior to participating in the auction. Show that $\beta$ remains an equilibrium even if resale is allowed. In particular, show that a bidder with value $x$ cannot gain by bidding an amount $b>\beta(x)$ even when he has the option of reselling the object to the other bidder.

Call the two bidders Jane and Mike. Given Mike's bid, and assuming that Mike follows the equilibrium strategy $\beta$, Jane can determine Mike's exact value. This implies that if Jane wins the item but realizes that she has lower value than Mike, she will resell it to Mike at exactly Mike's value.

We should calculate Jane's expected payoff when bidding $\beta(z)$ with $z\geq x$ when her value is $x$, and show that the expected payoff is maximized when $z=x$.

Jane wins with probability $F(z)$. If she loses, her payoff is $0$. If she wins and her value is higher than Mike's value, her payoff is $x-\beta(z)$. If she wins and her value is lower than Mike's value, $Y$, her payoff is $y-\beta(z)$. As a result, her expected payoff is $$-F(z)\beta(z)+F(x)x+(F(z)-F(x))E[Y\mid x<Y<z].$$

The term $F(x)x$ does not depend on $z$, so we are left to maximize $$-F(z)\beta(z)+(F(z)-F(x))E[Y\mid x<Y<z].$$

How do we maximize this?

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  • $\begingroup$ @Jamzy I've written in the post what I've tried so far. $\endgroup$
    – Alexi
    Commented Aug 30, 2015 at 22:51
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    $\begingroup$ You're right. I misread the unlighted parts as part of the question. $\endgroup$
    – Jamzy
    Commented Aug 30, 2015 at 22:53

2 Answers 2

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Let's look at this with backward induction. Let $v_{i}$ be my valuation and $v_{-i}$ be your valuation. Suppose I've won the item. Then $v_{i} \geq v_{-i}$. If I sell you the item at price $v_{-i}$, then my utility is $v_{-i} - b_{i} \leq v_{i} - b_{i}$, where $b_{i}$ is my bid according to the symmetric equilibrium bidding strategy (which may not necessarily be $\beta$). Notice that I do not gain $v_{i}$ by selling you the item, as I do not possess it.

We see that I will never have incentive to sell you the item. So I should bid as if I'm not going to sell you the item. We have an equilibrium bidding strategy for this anyways, given by the first price auction.

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    $\begingroup$ I edited and replaced $\beta(v_{i})$ with $b_{i}$. It's a force of habit to write $\beta(v_{i})$ when there is a symmetric equilibrium bidding strategy. I did not mean to imply the first price equilibrium bidding strategy. My apologies for the confusion. $\endgroup$
    – ml0105
    Commented Sep 9, 2015 at 21:29
  • $\begingroup$ Why does $v_i \geq v_{-i}$ follow from winning the item? I think the very point of the exercise is showing this. Only $b_i \geq b_{-i} = \beta(v_{-i})$ follows. $\endgroup$
    – Giskard
    Commented Sep 9, 2015 at 21:29
  • $\begingroup$ I edited a little. It's still a symmetric game, so there is a symmetric equilibrium bidding strategy which is non-decreasing on the valuations. It follows that if I win, your valuation cannot exceed mine. For if it did, then you would have bid at least as much as I did using the symmetric bidding strategy. $\endgroup$
    – ml0105
    Commented Sep 9, 2015 at 21:34
  • $\begingroup$ The way I understand the question: $(\beta,\beta)$ was a sym. eq. strategy profile in the old game. (no resale) Show that $(\beta,\beta)$ is also an eq. strat. profile in the new game. This implies that you have to show $$U_1(\beta',\beta) \leq U_1(\beta,\beta)$$. I think it is not enough to show that given $(\beta,\beta)$ no resale occurs. $\endgroup$
    – Giskard
    Commented Sep 9, 2015 at 21:37
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Let's first see why this problem is interesting: Suppose both players shade their bid according to the usual first price auction equilibrium. Whether these strategies are optimal crucially depends on what happens in the later subgames. Let's consider Jane who has 0 valuation for the good. In the usual first price auction, Jane should bid 0. However, this is now no longer true. In fact, Jane now has a positive valuation from winning the good in the auction, since she can sell it with some probability to Mike. Since the take-it-or-leave-it offer extracts the complete willingness to pay of Mike, Jane's valuation for the good is now $E[Y]$ where $Y$ is drawn from $F$. Similarly, all other types should behave as if they add to their valuation $y$ the amount $E[\bar Y|\bar Y>y]$ since this is the expected resale value of the good.

From this simple reasoning, it is straightforward to change the provided first price auction equilibrium condition

$\beta(x) = E[Y|Y<x]$

into a version which accounts for the increased valuation from resale:

$\beta(x) = E[Y+E[\bar Y|\bar Y > Y]\, | \, (Y+E[\bar Y|\bar Y > Y]) < (x + E[\bar Y|\bar Y > x]) ]$

where both $Y$ and $\bar Y$ are drawn from $F$.

Note 1: the equilibrium concept was not specified. The above solution assumes subgame perfection. The set of Nash equilibria is much larger, since offers can be rejected off the equilibrium path.

Note 2: since the strategies are one-to-one mappings of the drawn valuation, the players can always infer the valuation of the other player from the observed bid.

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  • $\begingroup$ So you are claiming that the exercise is mistaken and the $(\beta,\beta)$ from the first game is not a SPNE of the second game? I assume this because your provide a different formula for the equilibrium strategy of the second game, the 'second' $\beta$ . $\endgroup$
    – Giskard
    Commented Sep 10, 2015 at 15:19
  • $\begingroup$ indeed, i was a bit hasty with the conclusion that the new $\beta(x)$ differs from the previous one. it may still simplify to the original equilibrium. $\endgroup$
    – HRSE
    Commented Sep 13, 2015 at 2:10

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